Proving a function is continous only if a set is open Prove that a function from $\mathbb{R}$ to $\mathbb{R}$ is continuous if and only if the inverse image every open set is open.
Definition: A set is open if it contains all its interior points.
Proof: Let $f$ be continuous, then $\forall a,\forall \epsilon > 0$, there exists a $\delta > 0$ such that $$|x - a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon$$ Let $O$ be open to show $f^{-1}(O) = \{ x:f(x)\in O \}$ is open. Let $a\in f^{-1}(O) \Rightarrow f(a)\in O$. Since $O$ is open, there exists $\epsilon > 0$ such that $N(f(a),\epsilon)\subset O$. But, $f$ is continuous at $a$, so given this $\epsilon$,there exists a $\delta > 0$ such that $$|x - a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon$$ Moreover, $x\in N(a,\epsilon) \Rightarrow f(x)\in N(f(a),\epsilon)$. This last is contained in $O$, so  $N(a,\epsilon)\subset f^{-1}(O)$. Hence, $a$ is an interior point of $f^{-1}(O)$. This is true for all $a$, so $f^{-1}(O)$ is open.
I am not sure if this is right any suggestions would be greatly appreciated. 
 A: 
Moreover, $x\in N(a,\epsilon) \Rightarrow x\in N(f(a),\epsilon)$. This last is contained in $O$, so  $N(a,\epsilon)\subset f^{-1}(O)$. Hence, $a$ is an interior point of $f^{-1}(O)$. This is true for all $a$, so $f^{-1}(O)$ is open.

I suppose you mean $x\in N(a,\delta)\Rightarrow f(x)\in N(f(a),\epsilon)$. Then your proof is correct.
Also ...

Definition: A set is open if it contains all its interior points.

Here it should read "A set is open if each of its points is an interior point". I assume this is what you mean as it is exactly what you are showing in your proof.
What about the reverse implication? Or did you only need help for this direction?
A: Your definition of an open set is stated wrong; it is not possible for a set $O$ to have an interior point outside of $O$, so it doesn't make sense to define an open set as one that contains all its interior points. A closed set also contains all its interior points. Instead you want to say 

A set $O$ is open if every point of $O$ is an interior point. 

Or equivalently, 

A set $O$ is open if for every $x \in O$ there exists an open set $V$ containing $x$ such that $V \subset O$. 

Your proof looks on track until you say  "Moreover, $x\in N(a,\epsilon) \Rightarrow f(x)\in N(f(a),\epsilon)$. This last is contained in $O$, so  $N(a,\epsilon)\subset f^{-1}(O)$" because $x$ is in a $\delta$-ball, not an $\varepsilon$-ball. If you fix this to state that any point $a \in f^{-1}(O)$ can be contained in a $\delta$-ball that lies within $f^{-1}(O)$ then you are done.
